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Inverting Consequences of Uniform Convergence

  1. Jan 17, 2010 #1
    Hi. Now you probably know that if a function fk(x) converges uniformly to f(x) then we are allowed to certain actions such as

    [tex]limn-> \inf \int f (of k) dx = \int f dx [/tex]

    In other words we are allowed to exchange limit and integral. Now say we have any sequnce valued function fk(x) . And we also have

    [tex]limn-> \inf \int f (of k) dx = 0 [/tex]

    Does that imply that fk(x) converges uniformly to 0?
  2. jcsd
  3. Jan 17, 2010 #2


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    Please learn Latex properly.

    Indeed, if [tex](f_k)_{k\in\mathbb{N}}[/tex] converges uniformly to [tex]f[/tex] on some interval [tex][a,b]\subset\mathbb{R}[/tex], then [tex]\lim_{k\to\infty}\int_a^b f_k(x)dx=\int_a^b \lim_{k\to\infty}f_k(x)dx=\int_a^b f(x)dx[/tex].

    The converse of this is:

    if [tex]\lim_{k\to\infty}\int_a^b f_k(x)dx=\int_a^b f(x)dx[/tex], does it follow that [tex](f_k)_{k\in\mathbb{N}}[/tex] converges uniformly to [tex]f[/tex]?

    This is not what you asked, after all [tex]\int_a^b f(x)dx=0[/tex] does not imply [tex]f=0[/tex] (i.e. f(x)=0 for all x).

    The converse is therefore obviously not true. Take for example [a,b]=[0,1], and define f_k and f on [0,1] by f_k(x)=1/k and f(x)=-1 if x<1/2, f(x)=+1 if x>1/2. Then

    [tex]\lim_{k\to\infty}\int_0^1 f_k(x)dx=\lim_{k\to\infty}\frac{1}{k}=0=-\frac{1}{2}+\frac{1}{2}=\int_0^1 f(x)dx[/tex]

    but (f_k)_k does not even converge to f pointwise.

    For an counter-example to your claim, just take (f_k)_k to be a constant sequence: f_k=f as defined above, for all k. Then the integral of f_k is equal to zero (and hence the limit of the integral too), but f_k does not converge (even pointwise) to the zero function.
    Last edited: Jan 17, 2010
  4. Jan 17, 2010 #3
    Thanks a lot. And will do.
  5. Jan 17, 2010 #4


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    You're welcome. You can just click on any equation to see its Latex code.
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